# Back to basics: algebra and variables (nytimes and sony)

3 Jan

I’ve had many students get tripped up over the concept of variables, which is the key idea in algebra.  Unfortunately these were calculus students, so we had to try to catch up on algebra while learning calculus (also needed to review trigonometry and geometry.  No joke, once a student asked me how to find the area of a rectangle while I was proctoring a calculus exam.  I was speechless.  I checked back later and student had figured it out.)  But it’s not just calculus students who have trouble with algebra.  I ran into this tweet last week:

Snark aside, I want to solve the problem in this article.  First, let’s go way back to basics and talk about variables.  It’s absolutely a leap to go from the concrete world of numbers to something more representational.  Instead of jumping straight into abstraction, let’s use analogies!

First, note that variable implies something that varies.  While we used to use just numbers and equations like 8+2=10, now we have expressions like 8x+2 or x*x.  The value of this expression varies depending on what value we assign to x.  This is the tricky part, because we’re used to things that stand by themselves and don’t vary depending on some other context.  OR DO WE?

Let’s take this expression:

By itself, the sentence is ambiguous.  Are you saying you are buying extremely cheap cookies in Japan or Hawaii?  Or do you feel like eating some scrumptious small baked goods?  Or am I hanging out in your kitchen helping you bake?  In this expression, “yen” could mean three different things.  And depending on what value you put in for “yen” you’ll have completely different meanings.

How can we resolve the ambiguity?  Well, we could add some context.  For instance:

1. He offered me cake, but I had a yen for cookies.
2. I couldn’t afford a real meal, but I had a yen for cookies.

By adding context we narrowed down the sentence meaning to be unambiguous.  This is like adding an = sign to an expression.  So with 8x+2, the value of the expression could’ve been 10 (if x=1), or 0 (if x=-4), or 3000 (if x=374.75), or anything really.  But if we add an = sign and have 8x+2=18, then we’ve declared that 8x+2 is unambiguously 18.  So is unambiguously $\frac{18-2}{8} = 2$.

Sometimes we still might have multiple solutions to an equation.  My sentence could be “I was in Japan, and I had a yen for cookies.”  In fact, sentences like that basically make up all the captions in this blog.

So I could have an equation like x*x=1, and we’d have two solutions: x=1 works, and so does x=-1.

That’s that for the concept!  Let’s apply some algebra to that tweet.

We have a total of $15,000,000 brought in. Some of this is from$6 digital rentals, and some is from $15 sales. And we have 2000000 transactions in total. So let’s figure out how many transactions came from rentals, and how many from sales. Then we can figure out how much money sales brought in, and how much money rentals brought in. That was a heavy paragraph, so let’s break it down. What do we want in the end? How much money came from rentals, and how much came from sales. How can we calculate how much money came from rentals? Well, we can figure out how many rentals there were, and then multiply by 6, since each rental is$6.  Similarly, we can figure out the number of sales, and multiply by 15.  So let’s figure out the number of rentals and sales, using algebra.

First, we name our variables.  Let’s use for the number of rentals occurred, and use for the number of sales.  We know that R + S = 2,000,000 because there were two million transactions.  Also, we know that 6R + 15S= 15,000,000 because that’s how much money was brought in.  To make this a little easier to read, let’s cut out the 000,000 and add it in at the end.  Our system of equations is now:

• R+S=2
• 6R+15S=15

Now let’s solve.  If we solve the first equation for R, we’ll have R=2-S.  Notice that we have a variable in our solution for R, since the value of depends on the value of S.  But we can resolve the ambiguity by plugging this equation into our second equation.  Then we have

6(2-S)+15S=15

So, distributing, we have

12-6S+15S = 15

and grouping like terms gives

9S=15-12=3

Which means that S=3/9=1/3.  Going back to that solved equation earlier, this means that R=2-S=2-1/3=5/3.

So there were 1/3 million sales and 5/3 million rentals.  Going back to our “what are we looking for” paragraph, this means that $10,000,000 came from rentals, and$5,000,000 came from sales.

Of course, you could’ve just put this into wolfram instead of breaking it down into all the pieces like we did.  This post was mostly aimed at my old math 34a students at UCSB.  Likely the people who usually read the math posts on this blog won’t be interested in this post, but maybe you can pass it on to someone else!

### 7 Responses to “Back to basics: algebra and variables (nytimes and sony)”

1. Robert Pascual January 5, 2015 at 12:29 am #

One engineer I know got tripped up by the simple calculus question: if the derivative of y-squared is 2y then why is the derivative of 5-squared equal to zero?

• yenergy February 8, 2016 at 8:28 pm #

Somehow i missed this comment; I’m sorry! Thanks for stopping by. Yes, variables and constants get mixed up all the time. But also, this is hilarious!

2. royyman32 March 3, 2015 at 2:40 am #

Nice concise explanation. However especially with calculus students many of them do not go on to discover discrete math so their strengths probably rest elsewhere.

• yenergy February 8, 2016 at 8:30 pm #

Even if their strengths are somewhere else, and they “never” use math in their daily life, I think the lessons of quantitative reasoning and the necessity of context and data/information do help them forever